![Should I upgrade my town hall to 9? [closed]](https://edu.smbbmcl.edu.pk/uploads/should-i-upgrade-my-town-hall-to-9-closed_720.jpg)
Computing limit of $(1+1/n)^{n^2}$
By Sebastian Wright
How can I compute the limit $\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^{n^2}$? Of course $\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^{n} = e$, and then $\left(1+\frac{1}{n}\right)^{n^2} = \left(\left(1+\frac{1}{n}\right)^n\right)^n$. Since the term inside converges to $e$, the whole thing is $e^n$, and can I conclude that the limit is infinity as $n\rightarrow\infty$?
$\endgroup$ 33 Answers
$\begingroup$By Bernoulli's Inequality, $$\left(1+\frac{1}{n}\right)^{n^2}\geq 1+n$$ so the result follows trivially $\square$
$\endgroup$ 3 $\begingroup$There is a slight problem when you say "the whole things is $e^n$": what do you mean? Surely you know you cannot chose when to let the $n$s go to infinity, you have to make them large "at the same pace".
Take the logarithm of that, and see what happens. That is, you have $$x_n=\left(1+\frac 1 n\right)^{n^2}$$ Then $$\log x_n=n^2\log\left(1+\frac 1 n\right)$$
and $$\log\left(1+\frac 1n\right)=\frac 1n -\frac{1}{2n^2}+O\left(\frac 1{n^3}\right)$$
Alternatively, since $$\left(1+\frac 1 n\right)^n\to e $$
there exists $N$ such that whenever $n\geq N$, we have $$\left(1+\frac 1 n\right)^n\geq \frac{e}2\text{ (Why?) }$$
We have $\dfrac e2>1$ since $e>2$ and then $$x_n\geq r^n$$ whenever $n>N$ with $r>1$.
$\endgroup$ 0 $\begingroup$This question is ancient but I thought I'd give another easy way to solve this that basically only requires that you know $(1+\frac{1}{n})^n \to e$.
By $(1+\frac{1}{n})^n \to e>2$, there is an $N \in \mathbb{N}$ such that $n>N$ implies $(1+\frac{1}{n})^n > 2$. Then for $n> N$ we have $(1+\frac{1}{n})^{n^2} > 2^n$. For any large $M$, we thus have $(1+\frac{1}{n})^{n^2} > M$ for all $n> \max (N, \log_{2} M)$.
$\endgroup$
